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Area in square inches = (pi*radius^2)/2

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Q: What is the area of a 12 inch semicircle?
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Related questions

What is the equation to solve the area of a semicircle?

Area of a semicircle = (pi*radius^2)/2


Area of a 12 inch circle?

A 12-inch circle has an area of about 113.1 square inches.


What is the area of an 6 x 9 inch oval?

A 6 x 9 inch oval is a 3 inch x 3 inch square with a semicircle of 3 inch radius at each end. The total area is, therefore, 9 + 9*pi = 27.3 sq inches. Note that an oval is NOT an ellipse.


What is the area of the semicircle if the length of the arc of a semicircle is 10 and pi feet?

It is 27.49 units.


What is the area of a circle that has a 12 inch diameter?

The area of a circle that has a 12-inch diameter is: 113.1 square inches.


How many cubic feet in a 12 inch by 12 inch by 1 inch area?

one twelfth. 1/12


What is area of semicircle?

The area of a semicircle is half of the area of a circle. The area of a circle is pi x radius x radius. So the area of a semicircle is 1/2 x pi x radius x radius or (pi x radius x radius) / 2.


What is the area of a semicircle when the radius is 21?

The equation for area of a circle is A=pi x R x R. A semicircle is half a circle, so we calculate the area of a circle and divide by 2. The area of the circle would be approximately 1384.74. Thus the area of the semicircle would be 692.37.


How many 1 inch blocks in a 12 inch by 12 inch block?

Two measurements are indication of area. The area would be 144 square inches.


Area of 12 inch diameter circle vs 16 inch?

Area of a 12-inch circle: 113.1 square inches.Area of a 16-inch circle: 201.1 square inches.


What is the formula to find the area of a semicircle?

Pi*radius squared is how to find the area of a semicircle


What is the area outside the triangle if the triangle is inscribed in a semicircle while the side length of triangle is 2underroot2?

Diameter of semicircle = 1 Area of semicircle = Pi/8 Area of triangle = 0.25 Area outside triangle = (Pi/8) - 0.25 = 0.1427 (rounded to 4th decimal).